# Follow-up on $H_n(\mathbb{R}^3 - S^1)$

I'm trying to apply Mayer-Vietoris to compute $H_n(\mathbb{R}^3 - S^1)$ as asked by me here. Let $A := X -$"z axis" and $B:= B(0, 0.5) \times \mathbb{R}$ where $B(0, 0.5)$ is the open ball around $0$ with radius $0.5$.

The case $n=0$ is clear to me because $X$ is path connected so $H_0(X) = \mathbb{Z}$.

To compute $H_1(X)$ I wrote down the MV sequence as follows:

$$\dots H_1(A \cap B) \rightarrow H_1(A) \oplus H_1(B) \xrightarrow{i_\ast} H_1(X) \xrightarrow{\partial_\ast} H_0(A \cap B) \xrightarrow{j_\ast} H_0(A) \oplus H_0(B) \xrightarrow{g} H_0(X) \rightarrow 0$$

I know

$H_1(A \cap B) = \mathbb{Z}$,

$H_0(A) \oplus H_0(B) = H_1(A) \oplus H_1(B) = \mathbb{Z} \oplus \mathbb{Z}$ and

$H_0(A \cap B) = \mathbb{Z}$.

I also know that $j_\ast$ is injective so $ker j_\ast = 0 = im \partial_\ast \implies$ $\partial_\ast = 0 \implies$ $i_\ast$ is surjective

But here I think I'm stuck. What am I missing? Thanks for your help.

Edit Another question: it's not possible to compute $H_0$ using Mayer-Vietoris. I have to use that $X$ is path-connected. Right? Because the only information I can gain from MVS is that $g$ is surjective.

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Try to think about the map $H_1(A \cap B) \to H_1(A)$ geometrically. What cycles generate the homology of the torus? – Dylan Moreland Aug 16 '11 at 15:44
At some point you need further input. Isn't $A$ homeomorphic to $S^1 \times S^1 \times (0,1)$ ? Similarly, $A \cap B$ is homeomorphic to $S^1 \times (0,1) \times \mathbb R$. – Ryan Budney Aug 16 '11 at 18:10
@Dylan: the two cycles that generate $H_1(T^2)$ are the one that goes around the centre hole and the one that is perpendicular to it. Only the first is in $A \cap B$. From this I can conclude that $f: H_1(A \cap B) \rightarrow H_1(A)$ is not surjective. – Rudy the Reindeer Aug 18 '11 at 9:43
Ryan: I don't know what to make of this. – Rudy the Reindeer Aug 18 '11 at 9:50
@Dylan: and shouldn't I be looking at $H_1(A \cap B) \rightarrow H_1(A) \oplus H_1(B)$ instead? – Rudy the Reindeer Aug 18 '11 at 10:55

I used what I learned from this question here to compute this and I would appreciate it if you could tell me if my reasoning is correct:

$n=0$:

$$H_0(X) = \mathbb{Z}$$ because $X$ is path-connected.

$n=1$:

$$\dots \rightarrow H_1(A \cap B) \xrightarrow{(i,j)} H_1(A) \oplus H_1(B) \xrightarrow{k-l} H_1(X) \xrightarrow{\partial} H_0(A \cap B) \xrightarrow{(i^\prime , j^\prime)} H_0(A) \oplus H_0(B)\dots$$

Observations:

(i) $im (k-l) = H_1(X)$ because $ker (i^\prime , j^\prime) = 0 = im (\partial)$. $ker (i^\prime , j^\prime) = 0$ because $(i^\prime , j^\prime)((p,q)) = (0,0) \implies (p,q)=(0,0)$. $\implies k-l$ is surjective.

(ii) $ker (k-l) = im ((i,j)) \cong \mathbb{Z} \oplus 0 \cong \mathbb{Z}$ because the inclusion of $H_1(A) = H_1(S^1)$ maps the generator to a generator.

(iii) $im (k-l) = H_1(X) \cong (H_1(A) \oplus H_1(B))/ker(k-l) = (\mathbb{Z} \oplus \mathbb{Z})/ \mathbb{Z}$. Using the splitting lemma it follows that $H_1(X) \cong \mathbb{Z}$

$n=2$:

$$0 \xrightarrow{(i,j)} H_2(A) \oplus H_2(B) \xrightarrow{k-l} H_2(X) \xrightarrow{\partial} H_1(A \cap B) \xrightarrow{(i^\prime , j^\prime)} H_1(A) \oplus H_1(B) \dots$$

Observations:

(i) $im (k-l) = H_2(X)$ because $ker ((i^\prime , j^\prime)) = 0$ $\implies (k-l)$ is surjective.

(ii) $ker(k-l) = im((i,j)) = 0$ $\implies (k-l)$ is injective

(iii) $\implies H_2(X) \cong H_2(A) \oplus 0 = \mathbb{Z}$.

For $n = 1$, I don't understand your (i). Do you know what $H_1(X)$ is already? One way to see that $k - l$ is surjective is to note that the map $H_0(A \cup B) \to H_0(A) \oplus H_0(B)$ is injective. Careful with your parentheses on (iii), too! – Dylan Moreland Aug 21 '11 at 3:25
I have the analogous problem with your $n = 2$, (i) as well. Otherwise, nicely done! – Dylan Moreland Aug 21 '11 at 3:31